00:01
In this question, p -k of ethyl ammonium ion is given 10 .70 and we have to find out the ph of 0 .1 molar solution of ethylamine.
00:14
Ethylamine is a base that is ch3, ch2, nh2.
00:23
So for this, pkb will be calculated and ethyl ammonium ion is the acid.
00:33
Weak acid so that is nh3 positive and for it pka is given that is 10 .70 as the sum of the weak acid and its conjugate base is equal to 14 pk a plus pk b is b is equal to 14 bk b is equal to 14 pkb is can be calculated by 14 minus pca value is given 10 .70.
01:19
So this will be equals to 3 .3.
01:24
Now pkb is equals to minus log of kb.
01:31
So kb can be calculated as 10.
01:35
Point minus 3 .3 or it can be written as 5 into 10 to the power minus 4.
01:47
Now the ethylamine reacts with h2o and form ethyl ammonium ion, ch3, ch2, nh3 plus and oh negative ion.
02:10
At time t is equal to 0, the concentration of ethylamine and it is given 0 .1.
02:19
At time t0, say it's ethyl ammonium ion is 0 and oh is 0.
02:26
At time t is equals to equilibrium, c is equal to c minus alpha and it is approximately equals to c only...